Evaluate: `intcos^(-1)((1-x^2)/(1+x^2))\ dx`

Answer

To evaluate the integral \( I = \int \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \, dx \), we can follow these steps:

### Step 1: Substitution
Let \( x = \tan \theta \). Then, the differential \( dx \) becomes:
\[ dx = \sec^2 \theta \, d\theta \]

### Step 2: Rewrite the Integral
Substituting \( x = \tan \theta \) into the integral, we have:
\[ I = \int \cos^{-1} \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right) \sec^2 \theta \, d\theta \]
Using the identity \( \cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \), we can rewrite the integral as:
\[ I = \int \cos^{-1}(\cos 2\theta) \sec^2 \theta \, d\theta \]

### Step 3: Simplifying the Integral
Since \( \cos^{-1}(\cos 2\theta) = 2\theta \) (for \( 0 \leq 2\theta \leq \pi \)), we can simplify the integral:
\[ I = \int 2\theta \sec^2 \theta \, d\theta \]

### Step 4: Integration by Parts
Using integration by parts, let:
- \( u = 2\theta \) \(\Rightarrow du = 2 \, d\theta\)
- \( dv = \sec^2 \theta \, d\theta \) \(\Rightarrow v = \tan \theta\)

Now applying the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ I = 2\theta \tan \theta - \int \tan \theta (2) \, d\theta \]
\[ I = 2\theta \tan \theta - 2 \int \tan \theta \, d\theta \]

### Step 5: Integrating \( \tan \theta \)
The integral of \( \tan \theta \) is:
\[ \int \tan \theta \, d\theta = -\log |\cos \theta| + C \]
Thus:
\[ I = 2\theta \tan \theta + 2\log |\cos \theta| + C \]

### Step 6: Back Substitution
Now we need to substitute back \( \theta \) in terms of \( x \):
- From \( x = \tan \theta \), we have \( \theta = \tan^{-1}(x) \).
- Also, \( \tan \theta = x \) and \( \cos \theta = \frac{1}{\sqrt{1+x^2}} \).

Substituting these back into our expression for \( I \):
\[ I = 2 \tan^{-1}(x) \cdot x + 2 \log \left( \frac{1}{\sqrt{1+x^2}} \right) + C \]
\[ I = 2x \tan^{-1}(x) - \log(1+x^2) + C \]

### Final Answer
Thus, the evaluated integral is:
\[ \int \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \, dx = 2x \tan^{-1}(x) - \log(1+x^2) + C \]